Question

How do you find all the real and complex roots of `x^3 + 9x^2 + 19x - 29 = 0`?

Answer 1

`x=1,-5+-2i`

Explanation:

The potential real roots of a polynomial are the factors of `p/q`, where `p` is the constant and `q` is the coefficient of the term with the highest degree.

Since `p=29` and `q=1`, the possible real root(s) must be factors of `29/1=29`. This makes the problem significantly easier since the `29` is prime, so the only possibilities for the roots are `+-1,+-29`.

Use synthetic division or polynomial long division to find that `(x-1)` is a factor, which means that `color(red)1` is a root:

`(x^3+9x^2+19x-29)/(x-1)=x^2+10x+29`

The other two roots can be found through applying the quadratic formula on `x^2+10x+29`:

`x=(-b+-sqrt(b^2-4ac))/(2a)=(-10+-sqrt(-16))/2=color(red)(-5+-2i`